Getting derivative of the following function using L'Hopital?

[KataKoniK]

I have tried this question 5 times and each time I get a diff answer. I don't know what I am doing wrong, but here's the question. Any help?

lim
x-> 0

(4/x^2) - (2 / (1 - cos x))

edited - for typo in subject heading

[HallsofIvy]

First, you don't mean "getting the derivative", you mean "finding the limit".

Second, exactly what is the function? The way you have written it, I would interpret it as (4/x^2)- 2/(1- cos x). Since the numerators of both fractions are non-zero while the denominators go to 0, there is NO limit. Certainly L'Hopital would not apply at all here.

[KataKoniK]

Sorry about the typo.

Your interpretation is correct, so the answer DNE then? Furthermore, I just have a general question. How does 2sin x cosx−2x become sin 2x - 2x? I was just looking at past set solutions and it just suddenly does that. I don't understand

[Sirus]

That is due to the following trig identity:
\sin{2x}=2\sin{x}\cos{x}

[KataKoniK]

Got it. It's the double angle formula. Guess I just have to memorize it.

[Sirus]

That identity can be easily derived from the (hopefully) more familiar one below:
\sin{a+b}=\sin{a}\cos{b}+\sin{b}\cos{a}
Defining both a and b as the the same variable x, it is obvious that we get the previous identity for \sin{2x}. To be efficient at trigonometric limits, you should know these identities well, unless they are given to you in tests for reference.

[dextercioby]

I have tried this question 5 times and each time I get a diff answer. I don't know what I am doing wrong, but here's the question. Any help?limx-> 0(4/x^2) - (2 / (1 - cos x))

\lim_{x\rightarrow 0}\frac{4}{x^{2}}-\frac{2}{1-\cos x}=\lim_{x\rightarrow 0}\frac{4-4cosx-2x^{2}}{x^{2}(1-\cos x)}

=\lim_{x\rightarrow 0}\frac{4\sin x-4x}{2x(1-\cos x)+x^{2}\sin x}=
4\lim_{x\rightarrow 0}\frac{\cos x-1}{2(1-\cos x)+4x\sin x+x^{2}\cos x}


=-4\lim_{x\rightarrow 0}\frac{\sin x}{2\sin x+4\sin x+4x\cos x+2x\cos x-x^{2}\sin x}

=-4\lim_{x\rightarrow 0}\frac{\cos x}{6\cos x+6\cos x-8x\sin x-x^{2}\cos x}=-\frac{4}{12}=-\frac{1}{3}

Daniel.

[dextercioby]

Elegant way:
Use the fact that close to zero:
\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-...
,then the limit becomes:
\lim_{x\rightarrow 0}\frac{4}{x^{2}}-\frac{2}{1-1+\frac{x^{2}}{2}-\frac{x^{4}}{24}}=...=\lim_{x\rightarrow 0}\frac{48x^{2}-4x^{4}-48x^{2}}{12x^{4}-x^{6}}
=\lim_{x\rightarrow 0}\frac{-4}{12-x^{2}}=-\frac{1}{3}

Check it is still true for another term from the MacLaurin expansion of cosx
\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+...

Daniel.

[dextercioby]

Since the numerators of both fractions are non-zero while the denominators go to 0, there is NO limit.Certainly L'Hopital would not apply at all here.

Check post number 7.I guess i applied it 4 times. :wink:

Daniel.

[KataKoniK]

Thanks a lot Daniel. Saw my mistake!